Question

derivation for time period of satellite

Answer 1

Answer:

Time Period of Satellite:

⇒ For a satellite time taken to complete one round around the earth is known as time period of the satellite.

Derivation:

$\implies \sf T = \dfrac{Circumference\;of\;orbit}{Velocity}\\ \\ \\ \implies \sf T = \dfrac{2\pi r}{v_{o}}\\ \\ \\ \implies \sf T = \dfrac{2\pi r r^{1/2}}{\sqrt{GM}}\\ \\ \\ \implies \sf T = \dfrac{2\pi r^{3/2}}{\sqrt{GM}}\\ \\ \\ \implies \sf T=2\pi\sqrt{\dfrac{r^{3}}{GM}}\\ \\ \\ \sf As\;we\;know,\;r=Re+h\\ \\ \\ \implies {\boxed{\sf T= 2\pi \sqrt{\dfrac{(Re+h)^{3}}{GM}}}}\;\;\;\;.......(1)$

$\sf If\;h=0,\\ \\ \\ \implies \sf T=2\pi\sqrt{\dfrac{Re^{3}}{GM}}\;\;\;\;.....(2)\\ \\ \\ \sf As\;we\;know\;that,\; \rho=\dfrac{m}{v}\\ \\ \implies \sf m = v\times \rho \\ \\ \implies \sf m=\dfrac{4}{3}\pi Re^{3}\;.\rho\;\;\;\;......(3)\\ \\ \\ \sf Now,\;put\;equation\;(3)\;in\;equation\;(2)\;we\;get, \\ \\ \\ \implies \sf T=2\pi \sqrt{\dfrac{Re^{3}}{G\bigg(\dfrac{4}{3}\pi Re^{3}\rho\bigg)}}\\ \\ \\ \implies \sf T = 2\pi\sqrt{\dfrac{3}{G4\pi\rho}}$

$\implies \sf T=\sqrt{\dfrac{3\times 4\pi^{2}}{G4\pi\rho}}\\ \\ \\ \implies{\boxed{\sf T=\sqrt{\dfrac{3\pi}{G\rho}}}}$